Gm M m G. 2. The gravitational force between you and the moon at its initial position (directly opposite of Earth from you) is.

Transcription

1 Chapte 1 1 The avitational foce between the two pats is Gm M m G F = = mm m which we diffeentiate with espect to m and set equal to zeo: This leads to the esult m/m = 1/ df G = 0 = M m M = m dm The avitational foce between you and the moon at its initial position (diectly opposite of ath fom you) is mm F0 ( ) M whee M m is the mass of the moon, M is the distance between the moon and the ath, and is the adius of the ath At its final position (diectly above you), the avitational foce between you and the moon is F m m 1 ( M ) (a) The atio of the moon s avitational pulls at the two diffeent positions is 8 6 m /( M ) M 810 m 6710 m 8 6 m /( M ) M 810 m 6710 m F1 m F0 m Theefoe, the incease is , o appoximately 69% (b) The chane of the avitational pull may be appoximated as mm mm mm mm F1 F0 1 1 ( M ) ( M ) M M M M 4 m m M 618

2 619 On the othe hand, you weiht, as measued on a scale on ath, is F m m Since the moon pulls you up, the pecentae decease of weiht is m 7610 k 6710 m ( 10 )% 4 8 F M M k 810 m F F M THINK The manitude of avitational foce between two objects depends on thei distance of sepaation XPSS The manitude of the avitational foce of one paticle on the othe is iven by F = Gm 1 m /, whee m 1 and m ae the masses, is thei sepaation, and G is the univesal avitational constant ANALYZ Solve fo usin the values iven, we obtain Gm m F Nm / k 5k4k N LAN The foce of avitation is invesely popotional to 19m 4 We use subscipts s, e, and m fo the Sun, ath and Moon, espectively Pluin in the numeical values (say, fom Appendix C) we find 0 8 / sm s m sm s em k 810 m 4 11 em e m / em e sm k m F Gm m m 16 F Gm m m 5 The avitational foce fom ath on you (with mass m) is F m m whee mass If is the distance between you and a tiny black hole of b / 98 m/s k M that has the same avitational pull on you as the ath, then F m b m Combinin the two equations, we obtain

3 60 CHAPT 1 m m m 08 m b b (66710 m /k s )(110 k) 98 m/s 6 The avitational foces on m 5 fom the two 500 masses m 1 and m 4 cancel each othe Contibutions to the net foce on m 5 come fom the emainin two masses: F Nm /k 5010 k0010 k k net N 10 m The foce is diected alon the diaonal between m and m, towad m In unit-vecto notation, we have F net net ˆ ˆ 14 ˆ 14 ˆ F (cos 45i sin 45 j) ( N) i ( N) j 7 We equie the manitude of foce (iven by q 1-1) exeted by paticle C on A be equal to that exeted by B on A Thus, Gm A m C = Gm A m B d We substitute in m B = m A and m B = m A, and (afte cancelin m A ) solve fo We find = 5d Thus, paticle C is placed on the x axis, to the left of paticle A (so it is at a neative value of x), at x = 500d 8 Usin F = GmM/, we find that the topmost mass pulls upwad on the one at the oiin with N, and the ihtmost mass pulls ihtwad on the one at the oiin with N Thus, the (x, y) components of the net foce, which can be conveted to pola components (hee we use manitude-anle notation), ae F net 10410, (a) The manitude of the foce is N (b) The diection of the foce elative to the +x axis is THINK Both the Sun and the ath exet a avitational pull on the space pobe The net foce can be calculated by usin supeposition pinciple XPSS At the point whee the two foces balance, we have m/ 1 Sm/, whee M is the mass of ath, M S is the mass of the Sun, m is the mass of the space

4 61 pobe, 1 is the distance fom the cente of ath to the pobe, and is the distance fom the cente of the Sun to the pobe We substitute = d 1, whee d is the distance fom the cente of ath to the cente of the Sun, to find M M S d 1 1 ANALYZ Usin the values fo M, M S, and d iven in Appendix C, we take the positive squae oot of both sides to solve fo 1 A little aleba yields 11 d m MS / M 1 ( k)/( k) m LAN The fact that 1 Sun d indicates that the pobe is much close to the ath than the 10 Usin q 1-1, we find F AB = Gm A d j^, F AC = 4Gm A d i^ Since the vecto sum of all thee foces must be zeo, we find the thid foce (usin manitude-anle notation) is F AD = Gm A d (404 56º) This tells us immediately the diection of the vecto (pointin fom the oiin to paticle D), but to find its manitude we must solve (with m D = 4m A ) the followin equation: 404 Gm A d = Gm Am D This yields = 19d In manitude-anle notation, then, = (19 56º), with SI units undestood The exact answe without ead to sinificant fiue consideations is 6 6, (a) In (x, y) notation, the x coodinate is x = 0716d (b) Similaly, the y coodinate is y = 107d 11 (a) The distance between any of the sphees at the cones and the sphee at the cente is / cos0 /

5 6 CHAPT 1 whee is the lenth of one side of the equilateal tianle The net (downwad) contibution caused by the two bottom-most sphees (each of mass m) to the total foce on m 4 has manitude Gm4m Gm4m Fy = sin0 = This must equal the manitude of the pull fom M, so which eadily yields m = M Gm m Gm m ( / ) 4 4 (b) Since m 4 cancels in that last step, then the amount of mass in the cente sphee is not elevant to the poblem The net foce is still zeo 1 (a) We ae told the value of the foce when paticle C is emoved (that is, as its position x oes to infinity), which is a situation in which any foce caused by C vanishes (because q 1-1 has in the denominato) Thus, this situation only involves the foce exeted by A on B: GmAmB Fnet, x FAB N AB Since m B = 10 k and 00 m, then this yields AB m A F (00 m) ( N) 05 k ( m /k s )(10 k) 10 AB AB 11 GmB (b) We note (fom the aph) that the net foce on B is zeo when x = 040 m Thus, at that point, the foce exeted by C must have the same manitude (but opposite diection) as the foce exeted by A (which is the one discussed in pat (a)) Theefoe GmCm B = N m C = 100 k (040 m) 1 If the lead sphee wee not hollowed the manitude of the foce it exets on m would be F 1 = m/d Pat of this foce is due to mateial that is emoved We calculate the foce exeted on m by a sphee that just fills the cavity, at the position of the cavity, and subtact it fom the foce of the solid sphee The cavity has a adius = / The mateial that fills it has the same density (mass to volume atio) as the solid sphee, that is, M c / = M/, whee M c is the mass that fills the cavity The common facto 4/ has been canceled Thus,

6 6 M M c = M = M = 8 8 The cente of the cavity is d = d / fom m, so the foce it exets on m is The foce of the hollowed sphee on m is G M/8 m F = d / 1 1 m 1 F = F1 F = m = 1 d 8 d / d 81 /d 11 (66710 m /s k)(95 k)(041 k) N 1 (90010 m) 8[1 (410 m) /(910 m)] 14 All the foces ae bein evaluated at the oiin (since paticle A is thee), and all foces (except the net foce) ae alon the location vectos, which point to paticles B and C We note that the anle fo the location-vecto pointin to paticle B is 180º 00º = 150º (measued counteclockwise fom the +x axis) The component alon, say, the x axis of one of the foce vectos F is simply Fx/ in this situation (whee F is the manitude of F ) Since the foce itself (see q 1-1) is invesely popotional to, then the afoementioned x component would have the fom GmMx/ ; similaly fo the othe components With m A = k, m B = 0010 k, and m C = k, we theefoe have GmAmB xb GmAmC xc F net x = = ( N)cos(168º) B C and GmAmB yb GmAmC yc F net y = = ( N)sin(168º) B C whee B = d AB = 050 m, and (x B, y B ) = ( B cos(150º), B sin(150º)) (with SI units undestood) A faily quick way to solve fo C is to conside the vecto diffeence between the net foce and the foce exeted by A, and then employ the Pythaoean theoem This yields C = 040 m (a) By solvin the above equations, the x coodinate of paticle C is x C = 00 m (b) Similaly, the y coodinate of paticle C is y C = 05 m

7 64 CHAPT 1 15 All the foces ae bein evaluated at the oiin (since paticle A is thee), and all foces ae alon the location vectos, which point to paticles B, C, and D In thee dimensions, the Pythaoean theoem becomes = x + y + z The component alon, say, the x axis of one of the foce-vectos F is simply Fx/ in this situation (whee F is the manitude of F ) Since the foce itself (see q 1-1) is invesely popotional to then the afoementioned x component would have the fom GmMx/ ; similaly fo the othe components Fo example, the z component of the foce exeted on paticle A by paticle B is Gm A m B z B Gm A (m A )(d) = B ((d) + d + (d) ) = 4Gm A 7d In this way, each component can be witten as some multiple of Gm A /d Fo the z component of the foce exeted on paticle A by paticle C, that multiple is 9 14 /196 Fo the x components of the foces exeted on paticle A by paticles B and C, those multiples ae 4/7 and 14 /196, espectively And fo the y components of the foces exeted on paticle A by paticles B and C, those multiples ae /7 and 14 /98, espectively To find the distance to paticle D one method is to solve (usin the fact that the vecto add to zeo) Gm m Gm Gm d d A D A A 0449 With m D = 4m A, we obtain 1/ d 457d ( d ) 0449 The individual values of x, y, and z (locatin the paticle D) can then be found by considein each component of the Gm A m D / foce sepaately (a) The x component of would be which yields (b) Similaly, y = 90d, (c) and z = 0489d 4 14 A D A A Gm m x Gm Gm d d ma m (457 d) x d m d m d D A (4 A) In this way we ae able to deduce that (x, y, z) = (188d, 90d, 0489d),

8 65 16 Since the od is an extended object, we cannot apply quation 1-1 diectly to find the foce Instead, we conside a small diffeential element of the od, of mass dm of thickness d at a distance fom m 1 The avitational foce between dm and m1 is Gm1dm Gm1( M / L) d df, whee we have substituted dm ( M / L) d since mass is unifomly distibuted The diection of df is to the iht (see fiue) The total foce can be found by inteatin ove the entie lenth of the od: Gm1 M Ld d Gm1 M 1 1 Gm1 M F df d L L L d d d( L d) Substitutin the values iven in the poblem statement, we obtain 11 Gm1 M (66710 m /k s )(067 k)(50 k) F d ( L d ) (0 m)(0 m 0 m) N 17 (a) The avitational acceleation at the suface of the Moon is moon = 167 m/s (see Appendix C) The atio of weihts (fo a iven mass) is the atio of -values, so W moon = (100 N)(167/98) = 17 N (b) Fo the foce on that object caused by ath s avity to equal 17 N, then the fee-fall acceleation at its location must be a = 167 m/s Thus, a Gm Gm m a so the object would need to be a distance of / = 4 adii fom ath s cente 18 The fee-body diaam of the foce actin on the plumb line is shown to the iht The mass of the sphee is 4 4 M V (6 10 k/m )(00 10 m) k

9 66 CHAPT 1 The foce between the spheical mountain and the plumb line is F m / Suppose at equilibium the line makes an anle with the vetical and the net foce actin on the line is zeo Theefoe, m 0 Fnet, x T sin F T sin 0 F T cos m net, y The two equations can be combined to ive tan F m The distance the lowe end moves towad the sphee is x l tan l (050 m) m 11 1 (66710 m /k s )(87110 k) (98)(00 10 m) 19 THINK ath s avitational acceleation vaies with altitude XPSS The acceleation due to avity is iven by a = /, whee M is the mass of ath and is the distance fom ath s cente We substitute = + h, whee is the adius of ath and h is the altitude, to obtain a ( h) ANALYZ Solvin fo h, we obtain h / a Fom Appendix C, = m and M = k, so m / s k59810 k 49m / s h LAN We may ewite a as m 6 10 m a / h h (1 / ) (1 / ) whee 98 m/s is the avitational acceleation on the Suface of the ath The plot below depicts how a deceases with inceasin altitude

10 67 0 We follow the method shown in Sample Poblem 1 Diffeence in acceleation at head and feet Thus, a = da = d which implies that the chane in weiht is W W m da top bottom Howeve, since W bottom = GmM / (whee is ath s mean adius), we have GmM d m mda = d = W bottom = 600 N 00 N m fo the weiht chane (the minus sin indicatin that it is a decease in W) We ae not includin any effects due to the ath s otation (as teated in q 1-1) 1 Fom q 1-14, we see the exteme case is when becomes zeo, and pluin in q 1-15 leads to 0 = M = G Thus, with = 0000 m and = ad/s, we find M = k k (a) Pluin h = h /c into the indicated expession, we find a 4 h h c 1 = = = G M / c 00 h h h which yields a = ( k m/s ) /M h (b) Since M h is in the denominato of the above esult, a deceases as M h inceases (c) With M h = ( ) ( k), we obtain a = 98 m/s (d) This pat efes specifically to the vey lae black hole teated in the pevious pat With that mass fo M in q 1-16, and = 00/c, we obtain 6 c da = d = d 00 / c 00

11 68 CHAPT 1 whee d 170 m as in Sample Poblem 1 Diffeence in acceleation at head and feet This yields (in absolute value) an acceleation diffeence of m/s (e) The miniscule esult of the pevious pat implies that, in this case, any effects due to the diffeences of avitational foces on the body ae neliible (a) The avitational acceleation is a = = 76 m/s (b) Note that the total mass is 5M Thus, a G 5M = = 4 m/s 4 (a) What contibutes to the GmM/ foce on m is the (spheically distibuted) mass M contained within (whee is measued fom the cente of M) At point A we see that M 1 + M is at a smalle adius than = a and thus contibutes to the foce: F 1 on m G M M m (b) In the case = b, only M 1 is contained within that adius, so the foce on m becomes 1 m/b (c) If the paticle is at C, then no othe mass is at smalle adius and the avitational foce on it is zeo 5 Usin the fact that the volume of a sphee is 4 /, we find the density of the sphee: M a 4 total 1010 k m 4 10 k/m When the paticle of mass m (upon which the sphee, o pats of it, ae exetin a avitational foce) is at adius (measued fom the cente of the sphee), then whateve mass M is at a adius less than must contibute to the manitude of that foce (m/ ) (a) At = 15 m, all of M total is at a smalle adius and thus all contibutes to the foce: F GmM total 7 on m m 0 10 N/k (b) At = 050 m, the potion of the sphee at adius smalle than that is M 4 = =110 k

12 69 Thus, the foce on m has manitude m/ = m ( 10 7 N/k) (c) Pusuin the calculation of pat (b) alebaically, we find F 4 7 on m m Gm N k m 6 (a) Since the volume of a sphee is 4 /, the density is M M 4 4 total total When we test fo avitational acceleation (caused by the sphee, o by pats of it) at adius (measued fom the cente of the sphee), the mass M, which is at adius less than, is what contibutes to the eadin (/ ) Since M = (4 /) fo, then we can wite this esult as M total 4 G 4 total when we ae considein points on o inside the sphee Thus, the value a efeed to in the poblem is the case whee = : total a =, and we solve fo the case whee the acceleation equals a /: total total (b) Now we teat the case of an extenal test point Fo points with > the acceleation is total /, so the equiement that it equal a / leads to total total 7 (a) The manitude of the foce on a paticle with mass m at the suface of ath is iven by F = m/, whee M is the total mass of ath and is ath s adius The acceleation due to avity is

13 60 CHAPT 1 a F m m /s k59810 k m = = = = 98 m/s (b) Now a = /, whee M is the total mass contained in the coe and mantle toethe and is the oute adius of the mantle ( m, accodin to the fiue) The total mass is M = ( k k ) = k The fist tem is the mass of the coe and the second is the mass of the mantle Thus, a m /s k k m = = 984 m/s (c) A point 5 km below the suface is at the mantle cust inteface and is on the suface of a sphee with a adius of = m Since the mass is now assumed to be unifomly distibuted, the mass within this sphee can be found by multiplyin the mass pe unit volume by the volume of the sphee: M ( / e) Me, whee M e is the total mass of ath and e is the adius of ath Thus, m 4 4 M = k = k 6710 m The acceleation due to avity is a m /s k59110 k m = = = 979 m/s 8 (a) Usin q 1-1, we set GmM/ equal to 1 GmM/, and we find = Thus, the distance fom the suface is ( 1) = 0414 (b) Settin the density equal to M/V whee V = 4, we use q 1-19: 4 Gm 4 Gm M m 1 m F / 4 / 9 The equation immediately pecedin q 1-8 shows that K = U (with U evaluated at the planet s suface: J) is equied to escape Thus, K = J

14 61 0 The avitational potential eney is GmM m G U = = Mm m which we diffeentiate with espect to m and set equal to zeo (in ode to minimize) Thus, we find M m = 0, which leads to the atio m/m = 1/ to obtain the least potential eney Note that a second deivative of U with espect to m would lead to a positive esult eadless of the value of m, which means its aph is eveywhee concave upwad and thus its extemum is indeed a minimum 1 THINK Given the mass and diamete of Mas, we can calculate its mean density, avitational acceleation and escape speed XPSS The density of a unifom sphee is iven by = M/4, whee M is its mass and is its adius On the othe hand, the value of avitational acceleation a at the suface of a planet is iven by a = / fo a paticle of mass m, its escape speed is iven by 1 mm mv G v ANALYZ (a) Fom the definition of density above, we find the atio of the density of Mas to the density of ath to be M km = = 011 = km 4 M M M M (b) The value of avitational acceleation fo Mas is a M 4 M M M km a m/s 8 m/s M M M 4510 km M M M M (c) Fo Mas, the escape speed is v M 11 4 ( m /s k) k M m M m/s LAN The atio of the escape speeds on Mas and on ath is M / M M M 6510 km / M v M (011) 0455 v M km

15 6 CHAPT 1 (a) The avitational potential eney is U m m /s k5 k4 k 11 = = = J 19 m (b) Since the chane in potential eney is m m U = = J = 9 10 J, the wok done by the avitational foce is W = U = J (c) The wok done by you is W = U = J The amount of (kinetic) eney needed to escape is the same as the (absolute value of the) avitational potential eney at its oiinal position Thus, an object of mass m on a planet of mass M and adius needs K = GmM/ in ode to (baely) escape (a) Settin up the atio, we find Km Mm = = K M usin the values found in Appendix C m (b) Similaly, fo the Jupite escape eney (divided by that fo ath) we obtain KJ MJ = = 85 K M J 4 (a) The potential eney U at the suface is U s = J accodin to the aph, since U is invesely popotional to (see q 1-1), at an -value a facto of 5/4 times what it was at the suface then U must be 4 U s /5 Thus, at = 15 s, U = J Since mechanical eney is assumed to be conseved in this poblem, we have K + U = J at this point Since U = J hee, then 9 K 0 10 J at this point (b) To each the point whee the mechanical eney equals the potential eney (that is, whee U = J) means that U must educe (fom its value at = 15 s ) by a facto of, which means the value must incease (elative to = 15 s ) by a coespondin facto of Thus, the tunin point must be at = 5 s

16 6 5 Let m = 000 k and d = 0600 m (the oiinal ede-lenth, in tems of which the final ede-lenth is d/) The total initial avitational potential eney (usin q 1-1 and some elementay tionomety) is U i = 4Gm d Gm d Since U is invesely popotional to then educin the size by 1/ means inceasin the manitude of the potential eney by a facto of, so U f = U i U = U i = (4 + ) Gm d = J 6 ney consevation fo this situation may be expessed as follows: GmM GmM K U K U K K whee M = k, 1 = = m and m = 10 k (a) If K 1 = J and = m, then the above equation leads to 1 1 K K1 GmM J (b) In this case, we equie K = 0 and = m, and solve fo K 1 : 1 1 K1 K GmM J 7 (a) The wok done by you in movin the sphee of mass m B equals the chane in the potential eney of the thee-sphee system The initial potential eney is U and the final potential eney is Gm m Gm m Gm m d L L d A B A C B C i The wok done is U f Gm m Gm m Gm m L d L d A B A C B C

17 64 CHAPT W U f Ui GmB ma mc d L d L d d L d d L L d GmB ma mc GmB ( ma mc ) d( L d) d( L d) d( L d) 01 m (0040 m) 11 ( m / s k)(0010 k)(0080 k 000 k) (0040 m)( m) J (b) The wok done by the foce of avity is (U f U i ) = J 8 (a) The initial avitational potential eney is U i AM i B 11 ( m /s k)(0 k)(10 k) J J 080 m (b) We use consevation of eney (with K i = 0): 11 8 ( m /s k)(0 k)(10 k) Ui K U K 060 m which yields K = J Note that the value of is the diffeence between 080 m and 00 m 9 THINK The escape speed on the asteoid is elated to the avitational acceleation at the suface of the asteoid and its size XPSS We use the pinciple of consevation of eney Initially the paticle is at the suface of the asteoid and has potential eney U i = m/, whee M is the mass of the asteoid, is its adius, and m is the mass of the paticle bein fied upwad The initial kinetic eney is 1 mv The paticle just escapes if its kinetic eney is zeo when it is infinitely fa fom the asteoid The final potential and kinetic eneies ae both zeo Consevation of eney yields m/ + ½mv = 0 We eplace / with a, whee a is the acceleation due to avity at the suface Then, the eney equation becomes a + ½v = 0 Solvin fo v, we have v a

18 65 ANALYZ (a) Given that be 500 km and v a 0 m/s, we find the escape speed to a (0 m/s )( m) m/s (b) Initially the paticle is at the suface; the potential eney is U i = m/ and the kinetic eney is K i = ½mv Suppose the paticle is a distance h above the suface when it momentaily comes to est The final potential eney is U f = m/( + h) and the final kinetic eney is K f = 0 Consevation of eney yields m 1 m h mv We eplace with a and cancel m in the eney equation to obtain The solution fo h is 1 a ( h) a v a (0 m/s )( m) h a v (0 m/s )( m) (1000 m/s) m ( m) (c) Initially the paticle is a distance h above the suface and is at est Its potential eney is U i = m/( + h) and its initial kinetic eney is K i = 0 Just befoe it hits the asteoid its potential eney is U f = m/ Wite mv fo the final kinetic eney 1 f Consevation of eney yields m m 1 mv h We substitute a fo and cancel m, obtainin The solution fo v is a a h 1 v a (0 m/s )( m) v a (0 m/s )( m) h ( m) +( m) m/s LAN The key idea in this poblem is to ealize that eney is conseved in the pocess:

19 66 CHAPT 1 Ki Ui K f U f K U 0 The decease in potential eney is equal to the ain in kinetic eney, and vice vesa 40 (a) Fom q 1-8, we see that v0 / in this poblem Usin eney consevation, we have 1 mv 0 m/ = m/ which yields = 4 / So the multiple of is 4/ o 1 (b) Usin the equation in the textbook immediately pecedin q 1-8, we see that in this poblem we have K i = m/, and the above manipulation (usin eney consevation) in this case leads to = So the multiple of is 00 (c) Aain efein to the equation in the textbook immediately pecedin q 1-8, we see that the mechanical eney = 0 fo the escape condition 41 THINK The two neuton stas ae attacted towad each othe due to thei avitational inteaction XPSS The momentum of the two-sta system is conseved, and since the stas have the same mass, thei speeds and kinetic eneies ae the same We use the pinciple of consevation of eney The initial potential eney is U i = / i, whee M is the mass of eithe sta and i is thei initial cente-to-cente sepaation The initial kinetic eney is zeo since the stas ae at est The final potential eney is U /, whee the final sepaation is / We wite Mv fo the final kinetic eney of the system This f i is the sum of two tems, each of which is ½Mv Consevation of eney yields f f i i ANALYZ (a) The solution fo v is Mv 11 0 ( m / s k)(10 k) v m i m/s (b) Now the final sepaation of the centes is f = = 10 5 m, whee is the adius of eithe of the stas The final potential eney is iven by U f = / f and the eney equation becomes / i = / f + Mv The solution fo v is

20 v ( m / s k)(10 k) 5 10 f i 10 m 10 m m/s LAN The speed of the stas as a function of thei final sepaation is plotted below The decease in avitational potential eney is accompanied by an incease in kinetic eney, so that the total eney of the two-sta system emains conseved 4 (a) Applyin q 1-1 and the Pythaoean theoem leads to U = D + GmM y + D whee M is the mass of paticle B (also that of paticle C) and m is the mass of paticle A The value iven in the poblem statement (fo infinitely lae y, fo which the second tem above vanishes) detemines M, since D is iven Thus M = 050 k (b) We estimate (fom the aph) the y = 0 value to be U o = J Usin this, ou expession above detemines m We obtain m = 15 k 4 (a) If is the adius of the obit then the manitude of the avitational foce actin on the satellite is iven by m/, whee M is the mass of ath and m is the mass of the satellite The manitude of the acceleation of the satellite is iven by v /, whee v is its speed Newton s second law yields m/ = mv / Since the adius of ath is m, the obit adius is = ( m m) = m The solution fo v is 11 4 ( m / s k)( k) v m m/s (b) Since the cicumfeence of the cicula obit is, the peiod is T 6 (65 10 m) v m/s s This is equivalent to 875 min

21 68 CHAPT 1 44 Keple s law of peiods, expessed as a atio, is s s 1 s T T m Tm 1 luna month which yields T s = 05 luna month fo the peiod of the satellite 45 The peiod T and obit adius ae elated by the law of peiods: T = (4 /), whee M is the mass of Mas The peiod is 7 h 9 min, which is s We solve fo M: (94 10 m) M k GT 11 4 ( m / s k) s 46 Fom q 1-7, we obtain v = / fo the speed of an object in cicula obit (of adius ) aound a planet of mass M In this case, M = k and = ( )m = 7070 km = m The speed is found to be v = m/s Afte multiplyin by 600 s/h and dividin by 1000 m/km this becomes v = km/h (a) Fo a head-on collision, the elative speed of the two objects must be v = km/h (b) A pependicula collision is possible if one satellite is, say, obitin above the equato and the othe is followin a lonitudinal line In this case, the elative speed is iven by the Pythaoean theoem: = km/h 47 THINK The centipetal foce on the Sun is due to the avitational attaction between the Sun and the stas at the cente of the Galaxy XPSS Let N be the numbe of stas in the alaxy, M be the mass of the Sun, and be the adius of the alaxy The total mass in the alaxy is N M and the manitude of the avitational foce actin on the Sun is F ( NM ) GNM The foce, pointin towad the alactic cente, is the centipetal foce on the Sun Thus, Mv GNM F c F

22 69 The manitude of the Sun s acceleation is a = v /, whee v is its speed If T is the peiod of the Sun s motion aound the alactic cente then v = /T and a = 4 /T Newton s second law yields GNM / = 4 M/T The solution fo N is 4 N GT M ANALYZ The peiod is y, which is s, so N 0 4 ( 10 m) ( m / s k)( s) (0 10 k) LAN The numbe of stas in the Milky Way is between simplified model povides a easonable estimate 48 Keple s law of peiods, expessed as a atio, is to Ou a M T M T M (15) a T 1y whee we have substituted the mean-distance (fom Sun) atio fo the semi-majo axis atio This yields T M = 187 y The value in Appendix C (188 y) is quite close, and the small appaent discepancy is not sinificant, since a moe pecise value fo the semimajo axis atio is a M /a = 15, which does lead to T M = 188 y usin Keple s law A question can be aised eadin the use of a atio of mean distances fo the atio of semimajo axes, but this equies a moe lenthy discussion of what is meant by a mean distance than is appopiate hee 49 (a) The peiod of the comet is 140 yeas (and one month), which we convet to T = s Since the mass of the Sun is k, then Keple s law of peiods ives ( s) a a m 11 0 (66710 m /k s )( k) (b) Since the distance fom the focus (of an ellipse) to its cente is ea and the distance fom cente to the aphelion is a, then the comet is at a distance of ea a 1 1 (099 1) ( m) m when it is fathest fom the Sun To expess this in tems of Pluto s obital adius (found in Appendix C), we set up a atio:

23 640 CHAPT P 6,4 P 50 To hove above ath (M = k) means that it has a peiod of 4 hous (86400 s) By Keple s law of peiods, 4 7 (86400) m Its altitude is theefoe (whee = m), which yields m 51 THINK The satellite moves in an elliptical obit about ath An elliptical obit can be chaacteized by its semi-majo axis and eccenticity XPSS The eatest distance between the satellite and ath s cente (the apoee distance) and the least distance (peiee distance) ae, espectively, a = + d a = m m = m p = + d p = m m = m Hee = m is the adius of ath ANALYZ The semi-majo axis is iven by a p a m m m (b) The apoee and peiee distances ae elated to the eccenticity e by a = a(1 + e) and p = a(1 e) Add to obtain a + p = a and a = ( a + p )/ Subtact to obtain a p = ae Thus, 6 6 a p a p m m e a m m a p LAN Since e is vey small, the obit is nealy cicula On the othe hand, if e is close to unity, then the obit would be a lon, thin ellipse 5 (a) The distance fom the cente of an ellipse to a focus is ae whee a is the semimajo axis and e is the eccenticity Thus, the sepaation of the foci (in the case of ath s obit) is 11 9 ae m m (b) To expess this in tems of sola adii (see Appendix C), we set up a atio:

24 m m 5 Fom Keple s law of peiods (whee T = (4 h)(600 s/h) = 8640 s), we find the planet s mass M: (8640s) (8010 m) M k Howeve, we also know a = / = 80 m/s so that we ae able to solve fo the planet s adius: 11 4 (66710 m /k s )(40610 k) a 80 m/s m 54 The two stas ae in cicula obits, not about each othe, but about the two-sta system s cente of mass (denoted as O), which lies alon the line connectin the centes of the two stas The avitational foce between the stas povides the centipetal foce necessay to keep thei obits cicula Thus, fo the visible, Newton s second law ives Gm m m v F whee is the distance between the centes of the stas To find the elation between and 1, we locate the cente of mass elative to m 1 Usin quation 9-1, we obtain m (0) m m m m m1 m m1 m m On the othe hand, since the obital speed of m 1 is v 1 / T, then 1 vt / and the expession fo can be ewitten as m1 m vt m Substitutin and 1 into the foce equation, we obtain o 4 Gm m m v F ( m m ) v T T m vt (710 m/s) (170 days)(86400 s/day) ( m m ) G ( m /k s ) M, s k

25 64 CHAPT 1 0 whee M s k is the mass of the sun With m1 6M s, we wite m Ms and solve the followin cubic equation fo : (6 ) The equation has one eal solution: 9, which implies m / 9 M 55 (a) If we take the loaithm of Keple s law of peiods, we obtain 1 lo( T) = lo (4 / ) + lo ( a) lo ( a) lo ( T) lo (4 / ) whee we ae inoin an impotant subtlety about units (the auments of loaithms cannot have units, since they ae tanscendental functions) Althouh the poblem can be continued in this way, we pefe to set it up without units, which equies takin a atio If we divide Keple s law (applied to the Jupitemoon system, whee M is mass of Jupite) by the law applied to ath obitin the Sun (of mass M o ), we obtain s ( TT / ) = M a o M whee T = 655 days is ath s obital peiod and = m is its mean distance fom the Sun In this case, it is pefectly leitimate to take loaithms and obtain lo lo lo a T M T 1 Mo (witten to make each tem positive), which is the way we plot the data (lo ( /a) on the vetical axis and lo (T /T) on the hoizontal axis) (b) When we pefom a least-squaes fit to the data, we obtain

26 64 lo ( /a) = 0666 lo (T /T) + 101, which confims the expectation of slope = / based on the above equation (c) And the 101 intecept coesponds to the tem 1/ lo (M o /M), which implies Mo M M o 10 M Pluin in M o = k (see Appendix C), we obtain M = k fo Jupite s mass This is easonably consistent with the value k found in Appendix C 56 (a) The peiod is T = 7(600) = 9700 s, and we ae asked to assume that the obit is cicula (of adius = m) Keple s law of peiods povides us with an appoximation to the asteoid s mass: 4 16 (9700) M 6 10 k (b) Dividin the mass M by the iven volume yields an aveae density equal to = ( k)/( m ) = k/m, which is about 0% less dense than ath 57 In ou system, we have m 1 = m = M (the mass of ou Sun, k) With = 1 in this system (so 1 is one-half the ath-to-sun distance ), and v = /T fo the speed, we have T Gm1m m 1 T With = m, we obtain T = 10 7 s We can expess this in tems of athyeas, by settin up a atio: 7 T 10 s T (1y) = 7 1 y 071 y 1y s 58 (a) We make use of m vt ( m m ) G 1 whee m 1 = 09M Sun is the estimated mass of the sta With v = 70 m/s and T = 1500 days (o = s), we find

27 644 CHAPT 1 m (09 M m ) Sun k Since M Sun k, we find m k Dividin by the mass of Jupite (see Appendix C), we obtain m 7m J (b) Since v = 1 /T is the speed of the sta, we find 1 8 vt (70m/s)(1 10 s) m fo the sta s obital adius If is the distance between the sta and the planet, then = 1 is the obital adius of the planet, and is iven by m m m m m m Dividin this by m (ath s obital adius, ) ives = 5 59 ach sta is attacted towad each of the othe two by a foce of manitude /L, alon the line that joins the stas The net foce on each sta has manitude ( /L ) cos 0 and is diected towad the cente of the tianle This is a centipetal foce and keeps the stas on the same cicula obit if thei speeds ae appopiate If is the adius of the obit, Newton s second law yields ( /L ) cos 0 = Mv / The stas otate about thei cente of mass (maked by a cicled dot on the diaam above) at the intesection of the pependicula bisectos of the tianle sides, and the adius of the obit is the distance fom a sta to the cente of mass of the thee-sta system We take the coodinate system to be as shown in the diaam, with its oiin at the left-most sta The altitude of an equilateal tianle is / L, so the stas ae located at x = 0, y = 0; x = L, y = 0; and x = L/, y L/ The x coodinate of the cente of mass is x c = (L +

28 L/)/ = L/ and the y coodinate is c to the cente of mass is 645 y L / / L / The distance fom a sta x y L / 4 L /1 L / c c Once the substitution fo is made, Newton s second law then becomes / L cos0 Mv / L This can be simplified futhe by econizin that cos 0 / Divide the equation by M then ives /L = v /L, o v / L 60 (a) Fom q 1-40, we see that the eney of each satellite is m/ The total eney of the two satellites is twice that esult: 11 4 m (66710 m /ks )(59810 k)(15 k) A B m J (b) We note that the speed of the weckae will be zeo (immediately afte the collision), so it has no kinetic eney at that moment eplacin m with m in the potential eney expession, we theefoe find the total eney of the weckae at that instant is 11 4 ( m) (66710 m /ks )(59810 k)(15 k) J 6 (78710 m) (c) An object with zeo speed at that distance fom ath will simply fall towad the ath, its tajectoy bein towad the cente of the planet 61 The eney equied to aise a satellite of mass m to an altitude h (at est) is iven by U m h and the eney equied to put it in cicula obit once it is thee is ob Consequently, the eney diffeence is 1 m mv h, 1 1 m ( h ) (a) Solvin the above equation, the heiht h 0 at which 0 is iven by

29 646 CHAPT 1 1 ( h ) h m (b) Fo eate heiht h h0, 0, implyin 1 Thus, the eney of liftin is eate 6 Althouh altitudes ae iven, it is the obital adii that ente the equations Thus, A = ( ) km = 1740 km, and B = ( ) km = 5480 km (a) The atio of potential eneies is UB GmM / B A 1 U GmM / A A B (b) Usin q 1-8, the atio of kinetic eneies is KB GmM /B A 1 K GmM / A A B (c) Fom q 1-40, it is clea that the satellite with the laest value of has the smallest value of (since is in the denominato) And since the values of ae neative, then the smallest value of coesponds to the laest eney Thus, satellite B has the laest eney (d) The diffeence is GmM 1 1 B A B A Bein caeful to convet the values to metes, we obtain = J The mass M of ath is found in Appendix C 6 THINK We apply Keple s laws to analyze the motion of the asteoid XPSS We use the law of peiods: T = (4 /), whee M is the mass of the Sun ( k) and is the adius of the obit On the othe hand, the kinetic eney of any asteoid o planet in a cicula obit of adius is iven by K = m/, whee m is the mass of the asteoid o planet We note that it is popotional to m and invesely popotional to ANALYZ (a) The adius of the obit is twice the adius of ath s obit: = S = ( m) = m Thus, the peiod of the asteoid is

30 647 T (00 10 m) 11 0 ( m / s k)( k) s Dividin by (65 d/y) (4 h/d) (60 min/h) (60 s/min), we obtain T = 8 y (b) The atio of the kinetic eney of the asteoid to the kinetic eney of ath is K m /( ) m S 1 (010 ) 1010 K M /( ) M S 4 4 LAN An altenative way to calculate the atio of kinetic eneies is to use K mv / and note that v / T This ives K mv / m v m / T m T / S / S K M v M v M T M T 10 y (010 ) y in aeement with what we found in (b) (a) Cicula motion equies that the foce in Newton s second law povide the necessay centipetal acceleation: GmM v m Since the left-hand side of this equation is the foce iven as 80 N, then we can solve fo the combination mv by multiplyin both sides by = m Thus, mv = ( m) (80 N) = J Theefoe, K 1 mv J J (b) Since the avitational foce is invesely popotional to the squae of the adius, then Thus, F = (80 N) (/) = 6 N F F 65 (a) Fom Keple s law of peiods, we see that T is popotional to / (b) quation 1-8 shows that K is invesely popotional to

31 648 CHAPT 1 (c) and (d) Fom the pevious pat, knowin that K is popotional to v, we find that v is popotional to 1/ Thus, by q 1-1, the anula momentum (which depends on the poduct v) is popotional to / = 66 (a) The pellets will have the same speed v but opposite diection of motion, so the elative speed between the pellets and satellite is v eplacin v with v in q 1-8 is equivalent to multiplyin it by a facto of 4 Thus, K 11 4 ( m / k s ) k k m el 4 ( ) 10 m J (b) We set up the atio of kinetic eneies: K K el 1 bullet J k 950 m/s (a) The foce actin on the satellite has manitude m/, whee M is the mass of ath, m is the mass of the satellite, and is the adius of the obit The foce points towad the cente of the obit Since the acceleation of the satellite is v /, whee v is its speed, Newton s second law yields m/ = mv / and the speed is iven by v = / The adius of the obit is the sum of ath s adius and the altitude of the satellite: = ( ) m = m Thus, 11 4 ( m / s k) k v m (b) The peiod is m/s T = /v = ( m)/( m/s) = s 97 min (c) If 0 is the initial eney then the eney afte n obits is = 0 nc, whee C = J/obit Fo a cicula obit the eney and obit adius ae elated by = m/, so the adius afte n obits is iven by = m/ The initial eney is 11 4 ( m / s k) k 0 k ( m) 0 6 the eney afte 1500 obits is J,

32 649 nc J 1500 obit J obit J, and the obit adius afte 1500 obits is 11 4 ( m / s k) k 0 k 9 ( J) m The altitude is h = = ( m m) = m Hee is the adius of ath This toque is intenal to the satelliteath system, so the anula momentum of that system is conseved (d) The speed is 11 4 ( m / s k) k v m m / s 77 km/s (e) The peiod is T 6 ( m) s v m/s 9 min (f) Let F be the manitude of the aveae foce and s be the distance taveled by the satellite Then, the wok done by the foce is W = Fs This is the chane in eney: Fs = Thus, F = /s We evaluate this expession fo the fist obit Fo a complete obit s = = ( m) = m, and = J Thus, F J 7 s m 10 N () The esistive foce exets a toque on the satellite, so its anula momentum is not conseved (h) The satelliteath system is essentially isolated, so its momentum is vey nealy conseved 68 The obital adius is h km 400 km 6770 km m (a) Usin Keple s law iven in q 1-4, we find the peiod of the ships to be

33 650 CHAPT ( m) T ( m / s k)( k) s 9 min (b) The speed of the ships is 6 ( m) v m/s T s 0 (c) The new kinetic eney is K mv m(099 v0 ) (000 k)(099) (76810 m/s) J (d) Immediately afte the bust, the potential eney is the same as it was befoe the bust Theefoe, 11 4 m ( m / s k)( k)(000 k) 11 U J m (e) In the new elliptical obit, the total eney is K U J ( J) J (f) Fo elliptical obit, the total eney can be witten as (see q 1-4) m /a, whee a is the semi-majo axis Thus, m a 11 4 ( m / s k)( k)(000 k) 10 ( J) m () To find the peiod, we use q 1-4 but eplace with a The esult is 6 4 a 4 (66 10 m) T 11 4 ( m / s k)( k) s 895 min (h) The obital peiod T fo Picad s elliptical obit is shote than Io s by T T0 T 5540 s 570 s 170 s Thus, Picad will aive back at point P ahead of Io by 170 s 90 s = 80 s 69 We define the effective avity in his envionment as eff = 0/60 = 67 m/s Thus, usin equations fom Chapte (and selectin downwad as the positive diection), we find the fall-time to be

34 651 1 (1 m) y v0t eff t t 11 s 67 m/s 70 (a) The avitational acceleation a is defined in q 1-11 The poblem is concened with the diffeence between a evaluated at = 50 h and a evaluated at = 50 h + h (whee h is the estimate of you heiht) Assumin h is much smalle than 50 h then we can appoximate h as the d that is pesent when we conside the diffeential of q 1-11: da = d 50 h h = 50 (/c ) h If we appoximate da = 10 m/s and h 15 m, we can solve this fo M Givin ou esults in tems of the Sun s mass means dividin ou esult fo M by 10 0 k Thus, admittin some toleance into ou estimate of h we find the citical black hole mass should in the ane of 105 to 15 sola masses (b) Inteestinly, this tuns out to be lowe limit (which will supise many students) since the above expession shows da is invesely popotional to M It should pehaps be emphasized that a distance of 50 h fom a small black hole is much smalle than a distance of 50 h fom a lae black hole 71 (a) All points on the in ae the same distance ( = x + ) fom the paticle, so the avitational potential eney is simply U = m/ x +, fom q 1-1 The coespondin foce (by symmety) is expected to be alon the x axis, so we take a (neative) deivative of U (with espect to x) to obtain it (see q 8-0) The esult fo the manitude of the foce is mx(x + ) / (b) Usin ou expession fo U, the chane in potential eney as the paticle falls to the cente is 1 1 U m x By consevation of mechanical eney, this must tun into kinetic eney, K U mv / We solve fo the speed and obtain mv m v x x 7 (a) With 0 M 0 10 k and = m, we find a m/s (b) Althouh a close answe may be otten by usin the constant acceleation equations of Chapte, we show the moe eneal appoach (usin eney consevation):

35 65 CHAPT 1 K U K U o o whee K o = 0, K = ½mv, and U is iven by q 1-1 Thus, with o = m, we find 1 1 v o m/s 7 Usin eney consevation (and q 1-1) we have K 1 m 1 = K m (a) Pluin in two pais of values (fo (K 1, 1 ) and (K, )) fom the aph and usin the value of G and M (fo ath) iven in the book, we find m k (b) Similaly, v = (K/m) 1 / m/s (at = m) 74 We estimate the planet to have adius = 10 m To estimate the mass m of the planet, we equie its density equal that of ath (and use the fact that the volume of a sphee is 4 /): m M m M 4 / 4 / which yields (with M k and m) m = 10 7 k (a) With the above assumptions, the acceleation due to avity is a m /s k10 7 k Gm (10 m) m s 10 m s (b) quation 1-8 ives the escape speed: Gm v 00 m/s 75 We use m 1 fo the 0 k of the sphee at (x 1, y 1 ) = (05, 10) (SI units undestood), m fo the 40 k of the sphee at (x, y ) = (10, 10), and m fo the 60 k of the sphee at (x, y ) = (0, 05) The mass of the 0 k object at the oiin is simply denoted m We note that 1 15,, and = 05 (aain, with SI units undestood) The foce F n that the n th sphee exets on m has manitude Gmnm / n and is diected fom the oiin towad m n, so that it is conveniently witten as

36 65 Gmnm xn = ˆ y i + ˆ n Gmnm F j = ˆi + ˆ n j x n yn n n n n Consequently, the vecto addition to obtain the net foce on m becomes F F Gm m x m y n n ˆ n n ˆ 9 ˆ 7 ˆ net = n i j ( 9 10 N)i ( 10 N)j n=1 n1 n n1 n Theefoe, we find the net foce manitude is 7 F net 10 N 76 THINK We apply Newton s law of avitation to calculate the foce between the meteo and the satellite XPSS We use F = Gm s m m /, whee m s is the mass of the satellite, m m is the mass of the meteo, and is the distance between thei centes The distance between centes is = + d = 15 m + m = 18 m Hee is the adius of the satellite and d is the distance fom its suface to the cente of the meteo ANALYZ The avitational foce between the meteo and the satellite is F Gm m Nm / k 0k70k 18m s s N LAN The foce of avitation is invesely popotional to 77 We note that A (the distance fom the oiin to sphee A, which is the same as the sepaation between A and B) is 05, C = 08, and D = 04 (with SI units undestood) The foce F k that the k th sphee exets on m B has manitude Gmk mb / k and is diected fom the oiin towad m k so that it is conveniently witten as Gmk m B xk = ˆ y i + ˆ k Gmk mb F j = ˆi + ˆ k j x k yk k k k k Consequently, the vecto addition (whee k equals A, B, and D) to obtain the net foce on m B becomes mk x ˆ k mk y ˆ k 5 F ˆ net = Fk GmB i j (7 10 N)j k k k k k 78 (a) We note that C (the distance fom the oiin to sphee C, which is the same as the sepaation between C and B) is 08, D = 04, and the sepaation between sphees C and D is CD = 1 (with SI units undestood) The total potential eney is theefoe

37 654 CHAPT 1 M M M B C B D C D 4 = 1 10 J C D CD usin the mass-values iven in the pevious poblem (b) Since any avitational potential eney tem (of the sot consideed in this chapte) is necessaily neative (GmM/ whee all vaiables ae positive) then havin anothe mass to include in the computation can only lowe the esult (that is, make the esult moe neative) (c) The obsevation in the pevious pat implies that the wok I do in emovin sphee A (to obtain the case consideed in pat (a)) must lead to an incease in the system eney; thus, I do positive wok (d) To put sphee A back in, I do neative wok, since I am causin the system eney to become moe neative 79 THINK Since the obit is cicula, the net avitational foce on the smalle sta is equal to the centipetal foce XPSS The manitude of the net avitational foce on one of the smalle stas (of mass m) is m Gmm Gm m F M 4 This supplies the centipetal foce needed fo the motion of the sta: Gm m M m v 4 whee v / T Combinin the two expessions allows us to solve fo T ANALYZ Pluin in fo speed v, we aive at an equation fo the peiod T: T G( M m / 4) LAN In the limit whee m bodies M, we ecove the expected esult T fo two 80 If the anula velocity wee any eate, loose objects on the suface would not o aound with the planet but would tavel out into space

38 655 (a) The manitude of the avitational foce exeted by the planet on an object of mass m at its suface is iven by F = GmM /, whee M is the mass of the planet and is its adius Accodin to Newton s second law this must equal mv /, whee v is the speed of the object Thus, v = With M 4 / whee is the density of the planet, and v / T, whee T is the peiod of evolution, we find 4 4 G = T We solve fo T and obtain T G (b) The density is 0 10 k/m We evaluate the equation fo T: T m / s k010 k/m s 19 h 81 THINK In a two-sta system, the stas otate about thei common cente of mass XPSS The situation is depicted on the iht The avitational foce between the two stas (each havin a mass M) is F ( ) 4 The avitational foce between the stas povides the centipetal foce necessay to keep thei obits cicula Thus, witin the centipetal acceleation as whee is the anula speed, we have F F M 4 c ANALYZ (a) Substitutin the values iven, we find the common anula speed to be (66710 Nm /k )(0 10 k) 11 (1010 m) 7 10 ad/s

39 656 CHAPT 1 (b) To baely escape means to have total eney equal to zeo (see discussion pio to q 1-8) If m is the mass of the meteooid, then 1 GmM GmM 4 4 mv 0 v m/s LAN Compain with q 1-8, we see that the escape speed of the two-sta system is the same as that of a sta with mass M 8 The key point hee is that anula momentum is conseved: I p p = I a a which leads to ( / ),but p = a a whee a is detemined by q 1-4 p a p a (paticulaly, see the paaaph afte that equation in the textbook) Theefoe, p = a a ((T ) 1/ a ) = ad/s 8 THINK The obit of the shuttle oes fom cicula to elliptical afte chanin its speed by fiin the thustes XPSS We fist use the law of peiods: T = (4 /), whee M is the mass of the planet and is the adius of the obit Afte the obit of the shuttle tuns elliptical by fiin the thustes to educe its speed, the semi-majo axis is a m /, whee K U is the mechanical eney of the shuttle and its new peiod becomes T 4 a / ANALYZ (a) Usin Keple s law of peiods, we find the peiod to be T (4010 m) s 11 5 (66710 Nm /k )( k) (b) The speed is constant (befoe she fies the thustes), so v 7 (4010 m) 0 4 T 1510 s m/s (c) A two pecent eduction in the pevious value ives v v0 098(1 10 m/s) m/s

40 657 (d) The kinetic eney is 1 1 (000 k)( m/s) mv J K (e) Immediately afte the fiin, the potential eney is the same as it was befoe fiin the thuste: U 11 5 m (66710 N m /k )(95010 k)(000 k) m J (f) Addin these two esults ives the total mechanical eney: K U J ( 4510 J) 5 10 J () Usin q 1-4, we find the semi-majo axis to be m a 11 5 (66710 N m /k )(95010 k)(000 k) 11 ( 510 J) m (h) Usin Keple s law of peiods fo elliptical obits (usin a instead of ) we find the new peiod to be (40410 m) 4 T a 0 10 s 11 5 (66710 Nm /k )( k) This is smalle than ou esult fo pat (a) by T T = 1 10 s (i) Compain the esults in (a) and (h), we see that elliptical obit has a smalle peiod LAN The obits of the shuttle befoe and afte fiin the thuste ae shown below Point P coesponds to the location whee the thuste was fied 84 The diffeence between fee-fall acceleation and the avitational acceleation a at the equato of the sta is (see quation 114):